What is the shape of the `IBr_2^-` ion ?

Options

A. Linear
B. Bent shape with bond angle of about `90^@`
C. Bent shape with bond angle of about `109^@`
D. Bent shape with bond angle of about `120^@`

Answer

To determine the shape of the \( \text{IBr}_2^- \) ion, we can follow these steps:

### Step 1: Determine the Valence Electrons
- **Iodine (I)** has 7 valence electrons.
- Each **Bromine (Br)** has 7 valence electrons, and there are 2 Bromine atoms, contributing a total of \( 7 \times 2 = 14 \) electrons.
- The ion has an additional negative charge, which adds 1 more electron.

**Total valence electrons** = \( 7 + 14 + 1 = 22 \) electrons.

### Step 2: Draw the Lewis Structure
- Place the Iodine atom in the center, as it is less electronegative than Bromine.
- Connect the two Bromine atoms to the Iodine with single bonds. This uses up 4 electrons (2 for each bond).
- This leaves us with \( 22 - 4 = 18 \) electrons remaining to be distributed.

### Step 3: Distribute Remaining Electrons
- Place 6 electrons (3 lone pairs) around each Bromine atom to satisfy their octet.
- This uses up \( 6 \times 2 = 12 \) electrons, leaving us with \( 18 - 12 = 6 \) electrons.
- Place the remaining 6 electrons (3 lone pairs) on the Iodine atom.

### Step 4: Count Bonding and Lone Pairs
- **Bond pairs**: 2 (from the I-Br bonds)
- **Lone pairs**: 3 (on the Iodine atom)

### Step 5: Determine the Hybridization
- The total number of regions of electron density (bond pairs + lone pairs) around the Iodine atom is \( 2 + 3 = 5 \).
- This indicates that the hybridization is \( sp^3d \).

### Step 6: Determine the Molecular Geometry
- According to VSEPR theory, with 5 regions of electron density and 3 lone pairs, the arrangement will be **T-shaped**. The lone pairs will occupy the equatorial positions to minimize repulsion, while the bond pairs will occupy the axial positions.

### Conclusion
The shape of the \( \text{IBr}_2^- \) ion is **T-shaped**.

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